Improper Integrals: Type II and Related Concepts
Warm-up: Evaluating an Indefinite Integral with Partial Fractions and Limits
Original Integral: \int \frac{1}{(x-3)(x+1)}dx
Method: Partial Fraction Decomposition
Set up the decomposition: \frac{1}{(x-3)(x+1)} = \frac{A}{x-3} + \frac{B}{x+1}
Find common denominator: 1 = A(x+1) + B(x-3)
Solve for A and B:
Set x=3: 1 = A(3+1) + B(3-3) \implies 1 = 4A \implies A = \frac{1}{4}
Set x=-1: 1 = A(-1+1) + B(-1-3) \implies 1 = -4B \implies B = -\frac{1}{4}
Rewrite the Integral: \int \left( \frac{1}{4(x-3)} - \frac{1}{4(x+1)} \right) dx
Integrate:
= \frac{1}{4} \ln|x-3| - \frac{1}{4} \ln|x+1| + C
= \frac{1}{4} \ln \left| \frac{x-3}{x+1} \right| + C
Example Application to an Improper Integral (Implicit): The transcript hints at evaluating a definite integral that is improper, specifically where the upper bound approaches 3. If we consider \int_{0}^{3} \frac{1}{(x-3)(x+1)}dx, it would be defined as:
\lim{t \to 3^-} \int{0}^{t} \frac{1}{(x-3)(x+1)}dx
Evaluating the definite integral from 0 to t:
\left[ \frac{1}{4} \ln \left| \frac{x-3}{x+1} \right| \right]_{0}^{t} = \frac{1}{4} \ln \left| \frac{t-3}{t+1} \right| - \frac{1}{4} \ln \left| \frac{0-3}{0+1} \right|
= \frac{1}{4} \ln \left| \frac{t-3}{t+1} \right| - \frac{1}{4} \ln 3
Evaluate the limit as t \to 3^-:
As t \to 3^-, (t-3) approaches a small negative number from the left (i.e., (3-\epsilon)-3 = -\epsilon). But since it is inside an absolute value, it approaches a small positive number. (t+1) approaches 4.
Therefore, \lim_{t \to 3^-} \frac{1}{4} \ln \left| \frac{t-3}{t+1} \right| = \frac{1}{4} \ln \left( \frac{\text{small positive number}}{4} \right) = \frac{1}{4} ( -\infty) = -\infty
Conclusion: Since the limit is -\infty, this improper integral diverges.
L'Hôpital's Rule and Indeterminate Forms: The warm-up also briefly touches on indeterminate forms relevant to improper integrals.
An expression like \lim_{t \to 0^+} t \ln(t) is an indeterminate form of type 0 \cdot (- \infty).
To apply L'Hôpital's Rule, it must be rewritten as a fraction: \lim_{t \to 0^+} \frac{\ln(t)}{1/t}, which is of type \frac{-\infty}{\infty}.
Applying L'Hôpital's Rule: \lim{t \to 0^+} \frac{1/t}{-1/t^2} = \lim{t \to 0^+} (-t) = 0
Type II Improper Integrals: Infinite Discontinuity
Definition: Type II improper integrals occur when the integrand f(x) has an infinite discontinuity at one or both of the limits of integration (a or b) or at a point c within the interval [a,b].
An "infinite discontinuity" means that the function approaches \pm \infty at that point, making it undefined.
Case 1: Discontinuity at the Lower Bound (x=a)
Given \int_{a}^{b} f(x)dx, if f(x) is undefined at x=a, we rewrite the integral using a limit:
\int{a}^{b} f(x)dx = \lim{t \to a^+} \int_{t}^{b} f(x)dx
Case 2: Discontinuity at the Upper Bound (x=b)
Given \int_{a}^{b} f(x)dx, if f(x) is undefined at x=b, we rewrite the integral using a limit:
\int{a}^{b} f(x)dx = \lim{t \to b^-} \int_{a}^{t} f(x)dx
Case 3: Discontinuity within the Interval (x=c where a < c < b)
If f(x) is undefined at an interior point c, we split the integral into two parts:
\int{a}^{b} f(x)dx = \int{a}^{c} f(x)dx + \int_{c}^{b} f(x)dx
Each part is then handled as in Case 1 or Case 2: \lim{t \to c^-} \int{a}^{t} f(x)dx + \lim{u \to c^+} \int{u}^{b} f(x)dx
For the original integral to converge, BOTH resulting limits must converge.
Solving Procedure (Same as Type I):
Evaluate the definite integral normally, using t as one of the bounds.
Evaluate the limit of the result as t approaches the point of discontinuity.
If the limit exists as a finite number, the integral converges. Otherwise, it diverges.
Example: Evaluating \int_{0}^{1} \ln xdx
Identify Discontinuity: The function f(x) = \ln x is undefined as x \to 0^+ (where it approaches - \infty). This is a discontinuity at the lower bound (a=0). Therefore, we use Case 1.
Rewrite with a Limit:
\int{0}^{1} \ln xdx = \lim{t \to 0^+} \int_{t}^{1} \ln xdx
Evaluate the Indefinite Integral (\int \ln xdx):
Use Integration by Parts: \int u dv = uv - \int v du
Let u = \ln x \implies du = \frac{1}{x} dx
Let dv = dx \implies v = x
\int \ln xdx = (\ln x)(x) - \int (x)\left( \frac{1}{x} \right) dx
= x \ln x - \int 1 dx
= x \ln x - x + C
Evaluate the Definite Integral from t to 1:
\left[ x \ln x - x \right]_{t}^{1} = (1 \ln 1 - 1) - (t \ln t - t)
Since \ln 1 = 0: = (0 - 1) - (t \ln t - t)
= -1 - t \ln t + t
Evaluate the Limit:
\lim_{t \to 0^+} (-1 - t \ln t + t)
Separate the terms: \lim{t \to 0^+} (-1) - \lim{t \to 0^+} (t \ln t) + \lim_{t \to 0^+} (t)
We know that \lim_{t \to 0^+} (-1) = -1
We know from L'Hôpital's Rule (as demonstrated in the warm-up) that \lim_{t \to 0^+} (t \ln t) = 0
We know that \lim_{t \to 0^+} (t) = 0
Substitute these values: = -1 - 0 + 0 = -1
Conclusion: Since the limit exists and is a finite number (-1), the improper integral \int_{0}^{1} \ln xdx converges to -1.